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Absolute primary decomposition: bug?
https://www.singular.uni-kl.de/forum/viewtopic.php?f=10&t=2250		 Page 1 of 1
Author:  someuser [ Wed Apr 10, 2013 12:24 pm ]
Post subject:  Absolute primary decomposition: bug?
Using Singular 3.1.6 I tried to compute the absolute primary decomposition of the following ideal, defined over the rationals:

Code:
ring R = 0,T(1..4),dp;
ideal I =
T(3)^2+T(4)^2,
T(1)^2-T(2)^2-T(4)^2;

LIB "primdec.lib";

def S = absPrimdecGTZ(I);
setring S;
absolute_primes;


This results in the following output (to my knowledge the algorithm guesses random smooth points, so you may get different but isomorphic results):

Code:
[1]:
   [1]:
      _[1]=a^2+1
      _[2]=-47*T(1)*T(3)+T(3)^2-47*T(1)*T(4)*a+T(3)*T(4)*a
      _[3]=-2209*T(1)^2+2209*T(2)^2+T(3)^2+2210*T(4)^2
      _[4]=T(1)^2-T(2)^2-T(4)^2
   [2]:
      2


This means that there are two conjugate components, where for one of them you use a = i and for the other a = -i where i is the imaginary unit.

However, I think that these ideals are not prime (apply the above algorithm again on one of these components), neither are their radicals.

Using another software system, over QQ(i), i.e. the rationals with the imaginary unit adjoint, I get the following primary decomposition of I, which I think should be the result:

Code:
        T[1]^2 - T[2]^2 - T[4]^2,
        T[3] + K.1*T[4]


and

Code:
        T[1]^2 - T[2]^2 - T[4]^2,
        T[3] - K.1*T[4]


Is there a bug or did I use something in the wrong way?
Thanks in advance for your help.
Author:  someuser [ Wed Apr 10, 2013 12:26 pm ]
Post subject:  Re: Absolute primary decomposition: bug?
I forgot to say that K.1 just means the imaginary unit i.
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